How do you find the equation of the quadratic function that passes through the points #(1, 2)#, #(9, -4)# and #(11, -7)#?

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Answer 1 · 2018-05-05T15:36:12

Equation is #40y=-3x^2+83#

Let the equation be #y=ax^2+bx+c#. As it passes through #(1,2)#, #(9,-4)# and #(11,-7)#, we have

#a+b+c=2# .........................(A)

#81a+9b+c=-4# .........................(B)

and #121a+11b+c=-7# .........................(C)

Subtracting (A) from (B), we get #80a+8b=-6# ....................(D)

and subtracting (B) from (C), we get #40a+2b=-3# .................(E)

Now double (E) and subtract from (D) and we get

#4b=0# i.e. #b=0# and putting it in (E), we get #40a=-3# or #a=-3/40#

and now putting values of #a# and#b# in (A), we get

#c=2+3/40=83/40#

Hence quadratic equation is

#y=-3/40x^2+83/40# or #40y=-3x^2+83#

graph{(40y+3x^2-83)((x-1)^2+(y-2)^2-0.1)((x-9)^2+(y+4)^2-0.1)((x-11)^2+(y+7)^2-0.1)=0 [-16.25, 23.75, -11.92, 8.08]}